mathematical-analysis-prove-that-0-sin-3-x-x-3-dx-3pi-8-

Question Number 132987 by mnjuly1970 last updated on 17/Feb/21

....mathematical analysis... prove that:: 𝛗=∫_0 ^( ∞) ((sin^3 (x))/x^3 )dx=((3π)/8) ∗∗∗∗..........

$$\:\:\:\:\:\:\:\:\:\:\:\:\:….{mathematical}\:\:{analysis}… \\ $$$$\:{prove}\:\:{that}::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{3}} }{dx}=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\ast\ast\ast\ast………. \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 17/Feb/21

∫_0 ^∞ ((sin^3 x)/x^3 )dx =(3/4)∫_0 ^∞ ((sinx)/x^3 )−(1/4)∫_0 ^∞ ((sin3x)/x^3 )dx =(3/4).(π/(4sin(((3π)/2))))−(9/4)∫_0 ^∞ ((sinu)/u^3 )du ∫_0 ^∞ ((sinx)/x^μ )du=(π/(2Γ(μ)sin((π/2)μ))) =−(9/(4.4sin(((3π)/2))))−((3π)/(16))=((9π)/(16))−((3π)/(16))=((3π)/8)

$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}^{\mathrm{3}} {x}}{{x}^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\mathrm{3}{x}}{{x}^{\mathrm{3}} }{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}.\frac{\pi}{\mathrm{4}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}−\frac{\mathrm{9}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{sinu}}{{u}^{\mathrm{3}} }{du}\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{\mu} }{du}=\frac{\pi}{\mathrm{2}\Gamma\left(\mu\right){sin}\left(\frac{\pi}{\mathrm{2}}\mu\right)} \\ $$$$=−\frac{\mathrm{9}}{\mathrm{4}.\mathrm{4}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}−\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{9}\pi}{\mathrm{16}}−\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$

Commented by mnjuly1970 last updated on 18/Feb/21

$${thanks}\:{alot}\:{mr}\:{power}… \\ $$

Commented by Dwaipayan Shikari last updated on 18/Feb/21

$${Mr}\:{power}?? \\ $$$${hahaaa}….{sir} \\ $$

Commented by mnjuly1970 last updated on 18/Feb/21

$$\:{sorry}\:\:{sorry}… \\ $$$$\:\:{master}\:{Dwaipayan}.. \\ $$$$\:\: \\ $$

Answered by mnjuly1970 last updated on 18/Feb/21

another solution... ϕ(a)=∫_0 ^( ∞) ((sin^3 (ax))/x^3 )dx 𝛗=ϕ(1) ... ϕ′(a)=(∂/∂a)(∫_0 ^( ∞) ((sin^3 (ax))/x^3 )dx) =∫_0 ^( ∞) ((3xcos(ax)sin^2 (ax))/x^3 )dx =3∫_0 ^( ∞) ((sin^2 (ax)cos(ax))/x^2 ) dx =(3/2)∫_0 ^( ∞) ((sin(2ax)sin(ax))/x^2 )dx =(3/4)∫_0 ^( ∞) ((cos(ax)−cos(3ax))/x^2 )dx ϕ′′(a)=(3/4)∫_0 ^( ∞) ((−sin(ax))/x)dx+(9/4)∫_0 ^( ∞) ((sin(3ax))/x)dx =((3π)/4) ⇒ ϕ′(a)=(3/4)πa+C ϕ′(0)=0=C ϕ′(a)=(3/4)πa ⇒ ϕ(a)=(3/8)πa^2 +K ϕ(0)=0=K ϕ(a)=(3/8)πa^2 ⇒𝛗=ϕ(1)=(3/8)π ✓✓

$$\:\:\:{another}\:\:{solution}… \\ $$$$\:\:\:\varphi\left({a}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:\boldsymbol{\phi}=\varphi\left(\mathrm{1}\right)\:… \\ $$$$\:\:\:\:\varphi'\left({a}\right)=\frac{\partial}{\partial{a}}\left(\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{3}{xcos}\left({ax}\right){sin}^{\mathrm{2}} \left({ax}\right)}{{x}^{\mathrm{3}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{2}} \left({ax}\right){cos}\left({ax}\right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}{ax}\right){sin}\left({ax}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({ax}\right)−{cos}\left(\mathrm{3}{ax}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\varphi''\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{−{sin}\left({ax}\right)}{{x}}{dx}+\frac{\mathrm{9}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{3}{ax}\right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:\:\Rightarrow\:\varphi'\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\pi{a}+{C} \\ $$$$\:\:\:\:\varphi'\left(\mathrm{0}\right)=\mathrm{0}={C} \\ $$$$\:\:\:\varphi'\left({a}\right)=\frac{\mathrm{3}}{\mathrm{4}}\pi{a}\:\Rightarrow\:\varphi\left({a}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi{a}^{\mathrm{2}} +{K} \\ $$$$\:\:\:\:\varphi\left(\mathrm{0}\right)=\mathrm{0}={K} \\ $$$$\:\:\:\:\:\:\varphi\left({a}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi{a}^{\mathrm{2}} \:\:\Rightarrow\boldsymbol{\phi}=\varphi\left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{8}}\pi\:\checkmark\checkmark \\ $$

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